CHAPTER 14: ACIDS AND BASES

CHAPTER 14: ACIDS AND BASES Arrhenius Acids and Bases There are a few definitions of acids and bases, some are somewhat narrow and others are much broader. Arrhenius Acids dissociate when dissolved in water and produce hydrogen ions (H + ). Hydrogen chloride (HCl) is an Arrhenius acid. HCl(aq) H + (aq) + Cl (aq) Arrhenius Bases dissociate when dissolved in water and produce hydroxide ions (OH ). Sodium hydroxide (NaOH) is an Arrhenius base. NaOH(aq) Na + (aq) + OH (aq) Arrhenius Acids and Bases The Arrhenius definition correctly predicts the behavior of many acids and bases. However, this definition is limited and sometimes inaccurate. For example, H + does not exist in water. Instead, it reacts with water to form the hydronium ion, H 3 O +. H + (aq) + H 2 O(l) hydrogen ion: does not really exist in solution H 3 O + (aq) hydronium ion: actually present in aqueous solution Bronsted-Lowry Acids and Bases The Brønsted Lowry definition is more widely used: A Brønsted Lowry acid is a proton (H + ) donor. A Brønsted Lowry base is a proton (H + ) acceptor. This proton is donated. HCl(g) + H 2 O(l) H 3 O + (aq) + Cl (aq) HCl is a Brønsted Lowry acid because it donates a proton to the solvent water. H 2 O is a Brønsted Lowry base because it accepts a proton from HCl. A Brønsted Lowry acid must contain a hydrogen atom. Common Brønsted Lowry acids (HA): Acids Can Have Different # of H-Atoms A monoprotic acid contains one acidic proton. HCl HCl hydrochloric acid HBr hydrobromic acid HNO 3 nitric acid H H 2 SO 4 sulfuric acid H C C O H O H acetic acid acidic H atom A diprotic acid contains two acidic protons. H 2 SO 4 A triprotic acid contains three acidic protons. H 3 PO 4 A Brønsted Lowry acid may be neutral or it may carry a net positive or negative charge. HCl, H 3 O +, HSO 4 1

Bronsted-Lowry Bases Common Bronsted-Lowry Bases A Brønsted Lowry base is a proton acceptor, so it must be able to form a bond to a proton. A base must contain a lone pair of electrons that can be used to form a new bond to the proton. Lone pairs make these neutral compounds bases. NH 3 ammonia The OH is the base in each metal salt. NaOH sodium hydroxide This e pair forms a new bond to a H from H 2 O. H + H 2 O water KOH potassium hydroxide H N H + H 2 O(l) H N H Brønsted Lowry base H H + OH (aq) Mg(OH) 2 magnesium hydroxide Ca(OH) 2 calcium hydroxide The Reaction of a Bronsted-Lowry Acid With a Bronsted-Lowry Base This e pair stays on A. This e pair forms a new bond to H +. gain of H + H A + B A + H B + acid base loss of H + Bronsted-Lowry Conjugate Acid-Base Pairs gain of H + H A + B A + H B + acid base conjugate conjugate base acid loss of H + The product formed by loss of a proton from an acid is called its conjugate base. The product formed by gain of a proton by a base is called its conjugate acid. Bronsted-Lowry Conjugate Acid-Base Pairs Bronsted-Lowry Conjugate Acid-Base Pairs gain of H + H Br + H 2 O Br + H 3 O + acid base conjugate conjugate base acid loss of H + HBr and Br are a conjugate acid base pair. H 2 O and H 3 O + are a conjugate acid base pair. The net charge must be the same on both sides of the equation. 2

Bronsted-Lowry Conjugate Acid-Base Pairs There are 2 conjugate acid-base pairs in each reaction. HF(aq) + H 2 O(l) F (aq) + H 3 O + (aq) In this acid-base reaction Conjugate acid-base pairs are related to each other by H + HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 (aq) acid base Conjugate acid Conjugate base conjugate pair conjugate pair An acid, HF, donates H + to form its conjugate base, F A base, H 2 O, accepts H + to form its conjugate acid, H 3 O + There are two conjugate acid-base pairs (HF / F ) and (H 2 O / H 3 O + ), and each pair is related by an H + Identify each Bronsted-Lowry Acid and Base and the Conjugate Acid-Base Pairs OCl (aq) + H 2 O(l) HOCl(aq) + OH (aq) How to Write the Formulas of Conjugate Acids and Bases When a species gains a proton (H + ) to form a conjugate acid, it gains a +1 charge. H 2 C 2 O 4 (aq) + H 2 O(l) HPO 4 2 (aq) + H 2 O(l) H 3 O + (aq) + HC 2 O 4 (aq) H 3 O + (aq) + PO 4 3 (aq) H 2 O base zero charge add H + H 3 O + +1 charge When a species loses a proton (H + ) to form a conjugate base, it effectively gains a 1 charge. H 3 AsO 4 (aq) + H 2 O(l) H 3 O + (aq) + H 2 AsO 4 (aq) HBr acid zero charge lose H + Br 1 charge Write the formulas of the conjugate acids for the following Brønsted Lowry bases: Remember: Bases are substances that accept a proton (H + ). Write the formulas of the conjugate bases for the following Brønsted Lowry acids: Remember: Acids are substances that donate a proton (H + ). a. ClCH 2 CO 2 + H + ClCH 2 CO 2 H a. HCN H + CN b. C 5 H 5 N + H + C 5 H 5 NH + b. (CH 3 ) 2 NH + 2 H + (CH 3 ) 2 NH c. SeO 4 2 + H + HSeO 4 c. H 3 PO 4 H + H 2 PO 4 d. (CH 3 ) 3 N + H + (CH 3 ) 3 NH + d. HSeO 3 H + SeO 3 2 Remember: Add H + to write the conjugate acid. Remember: Remove H + to write the conjugate base. 3

Identifying Conjugate Pairs in a Bronsted Acid-Base Reaction What is (a) the conjugate base on HNO 3, (b) the conjugate acid of O 2-, (c) the conjugate base of HSO 4-, and (d) the conjugate acid of HCO 3-. Strategy To find the conjugate base of a species, remove a proton from the formula. To find the conjugate acid of a species, add a proton to the formula. The word proton, in this context, refers to H +. Thus, the formula and the charge will both be affected by the addition or removal of H +. Solution (a) NO 3 - (b) OH - (c) SO 4 2- (d) H 2 CO 3 Think About It A species does not need to be what we think of as an acid in order for it to have a conjugate base. For example, we would not refer to the hydroxide ion (OH - ) as an acid but it does have a conjugate base, the oxide ion (O 2- ). Furthermore, a species that can either lose or gain a proton, such as HCO 3-, has both a conjugate base (CO 3 2- ) and a conjugate acid (H 2 CO 3 ). Identifying Conjugate Pairs in a Bronsted Acid-Base Reaction Label each of the species in the following equations as an acid, base, conjugate base, or conjugate acid: (a) HF(aq) + NH 3 (aq) F - (aq) + NH 4+ (aq) (b) CH 3 COO - (aq) + H 2 O(l) CH 3 COOH(aq) + OH - (aq) Strategy Think In About each equation, It In a Brønsted the reactant acid-base that loses reaction, a proton there is the is always acid and the reactant an that acid gains and a the base, proton and is whether base. a substance Each product behaves is the as conjugate an acid or of a one of the reactants. base depends Two species on what that it differ is combined only by with. a proton Water, constitute for example, a conjugate pair. behaves as a base when combined with HCl but behaves as an acid Solution when (a) combined HF loses with a proton NH a becomes F - ; NH 3 gains a proton and becomes 3. NH 4+. HF(aq) + NH 3 (aq) F - (aq) + NH 4+ (aq) acid base conjugate base conjugate acid (b) CH 3 COO - gains a proton to become CH 3 COOH; H 2 O loses a proton to become OH -. CH 3 COO - (aq) + H 2 O(l) CH 3 COOH(aq) + OH - (aq) base acid conjugate acid conjugate base Amphoteric Compounds Can Behave as Either an Acid or a Base An Amphoteric compound contains both a H-atom and a lone pair of e and can behave either as an acid or a base. H O H H 2 O as a base H O H H 2 O as an acid H add H + H O H conjugate acid remove H + H O + conjugate base Two Sets of Rules for Naming Acids Set #1: Binary acids Hydrogen bonded to a nonmetal Examples: HCl, HBr, HI, H 2 S Set #2: Oxyacids Hydrogen bonded to a polyatomic ion Examples: H 2 SO 4, HNO 3, HClO 4, HCH 3 CO 2 Is the compound binary or not binary? Naming Binary Acids hydro - nonmetal root ic acid 1. The word hydrogen in the anhydrous compound name is dropped. 2. The prefix hydro- is attached to the stem of the name of the nonmetal that is combined with hydrogen. 3. The suffix -ide on the stem of the nonmetal that is combined with hydrogen is replaced with the suffix -ic. 4. The word acid is added the the end of the name as a separate word. 5. If the nonmetal is sulfur or phosphorus, the stems sulfand phosph- are expanded to sulfur- and phosphor- for pronunciation reasons before the -ic suffix is added. Naming Binary Acids HCl HBr hydrogen chloride hydrogen bromide hydro + chlor + ic acid hydro + brom + ic acid hydrochloric acid hydrobromic acid H 2 S hydrogen sulfide hydro + sulfur + ic acid hydrosulfuric acid 4

A compound must contain at least one ionizable hydrogen atom to be an acid upon dissolving. Naming Oxyacids 1. All hydrogens that are written as the first part of the formula of the acid are dropped. 2. The polyatomic ion that remains is named. 3. If the polyatomic ion ends with the suffix -ate, the suffix is replaced by the suffix -ic, and the word acid is added. 4. If the polyatomic ion ends with the suffix -ite, the suffix is replaced by the suffix -ous, and the word acid is added. 5. If the polyatomic ion contains sulfur or phosphorus, the stems sulf- and phos- are expanded to sulfur- and phosphor- for pronunciation reasons before the -ic or - ous suffixes are added. Naming Oxyacids Names and Formulas of Common Acids H 2 SO 4 SO 4 2 sulfate ion sulfuric acid HNO 3 NO 3 nitrate ion nitric acid HNO 2 NO 2 nitrite ion nitrous acid Naming Bases Bases with OH ions are named as the hydroxide of the metal in the formula. NaOH KOH Ba(OH) 2 Al(OH) 3 Fe(OH) 3 sodium hydroxide potassium hydroxide barium hydroxide aluminum hydroxide iron(iii) hydroxide Al(OH) 3 KOH HBr HNO 2 H 2 SO 4 HBrO 2 Name Each of the Following Acids and Bases Aluminum hydroxide Potassium hydroxide Hydrobromic acid Nitrous acid Sulfuric acid Bromous acid 5

Write Formulas for Each of the Following Acids and Bases Barium hydroxide Nitric acid Sodium hydroxide Hydroiodic acid Strontium hydroxide Ba(OH) 2 HNO 3 NaOH HI Sr(OH) 2 Comparing Acids and Bases Characteristic Acids Bases Arrhenius Produces H + in water Produces OH in water Bronsted-Lowry Donates H + Accepts H + Electrolytes Yes Yes Taste Sour Bitter Feel May sting Slippery Litmus Red Blue Phenolphthalein Colorless Pink Strength of Acids The strength of an acid is a measure of how completely it dissociates (ionizes) when dissolved. HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl (aq) Strong acid dissociations are not treated as equilibria, rather as processes that go to completion. Hydrochloric acid HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl (aq) When a strong acid dissolves in water, 100% of the acid dissociates into ions. Hydrobromic acid Hydroiodic acid Nitric acid HBr(aq) + H 2 O(l) HI(aq) + H 2 O(l) HNO 3 (aq) + H 2 O(l) H 3 O + (aq) + Br (aq) H 3 O + (aq) + I (aq) H 3 O + (aq) + NO 3 (aq) A single reaction arrow is used, because the product is greatly favored at equilibrium. Chloric acid Perchloric acid HClO 3 (aq) + H 2 O(l) HClO 4 (aq) + H 2 O(l) H 3 O + (aq) + ClO 3 (aq) H 3 O + (aq) + ClO 4 (aq) Common strong acids are HI, HBr, HCl, H 2 SO 4, and HNO 3. Sulfuric acid H 2 SO 4 (aq) + H 2 O(l) H 3 O + (aq) + HSO 4 (aq) Strength of Acids The strength of an acid is a measure of how completely it dissociates (ionizes) when dissolved. CH 3 COOH(l) + H 2 O(l) H 3 O + (aq) + CH 3 COO (aq) When a weak acid dissolves in water, only a small fraction of the acid dissociates into ions. Strength of Acids Unequal reaction arrows may be used, because the reactants are usually favored at equilibrium. Common weak acids are H 3 PO 4, HF, H 2 CO 3, and HCN. 6

Comparing Strong and Weak Acids 6 Strong Acids (Memorize) A strong acid, HCl, is completely dissociated into H 3 O + (aq) and Cl (aq). A weak acid, CH 3 COOH, contains mostly undissociated acid. HCl hydrochloric acid HBr hydrobromic acid HI hydroiodic acid HClO 4 perchloric acid HNO 3 nitric acid H 2 SO 4 sulfuric acid Strong acids have weak conjugate bases. Strength of Bases When a strong base dissolves in water, 100% of the base dissociates into ions. Comparing Strong and Weak Bases NaOH(s) + H 2 O(l) Na + (aq) + OH(aq) When a weak base dissolves in water, only a small fraction of the base dissociates into ions. NH 3 (g) + H 2 O(l) NH 4+ (aq) + OH(aq) A strong base, NaOH, is completely dissociated into Na + (aq) and OH(aq). A weak base contains mostly undissociated base, NH 3. Strong Bases are Hydroxides of Groups 1A and 2A Metals LiOH NaOH KOH RbOH CsOH lithium hydroxide sodium hydroxide potassium hydroxide rubidium hydroxide cesium hydroxide Ca(OH) 2 calcium hydroxide Sr(OH) 2 strontium hydroxide Ba(OH) 2 barium hydroxide Relating Acid and Base Strength A strong acid readily donates a proton, forming a weak conjugate base. HCl strong acid Cl weak conjugate base A strong base readily accepts a proton, forming a weak conjugate acid. OH strong base H 2 O weak conjugate acid 7

Strong acids have weak conjugate bases. The position of the equilibrium depends on the strengths of the acids and bases. (Competition for the Proton) H A + B A + H B + acid base conjugate base conjugate acid The stronger acid reacts with the stronger base to form the weaker acid and the weaker base. Weak acids have strong conjugate bases. Stronger acid + Stronger base Weaker acid + Weaker base The position of the equilibrium depends on the strengths of the acids and bases. When the stronger acid and base are the reactants on the left side, the reaction readily occurs and the reaction proceeds to the right. H A + B A + H B + stronger acid stronger base weaker base weaker acid If you mix equal concentrations of reactants and products, will the reaction proceed to the left or the right (where does the equilibrium lie)? H 2 SO 4 (aq) + NH 3 (aq) NH 4+ (aq) + HSO 4 (aq) acid base acid base H 2 SO 4 is a stronger acid than NH 4+, and NH 3 is a stronger base than HSO 4. Therefore, NH 3 gets the proton and the reaction proceeds to the right. H 2 SO 4 (aq) + NH 3 (aq) NH 4+ (aq) + HSO 4 (aq) A larger forward arrow means that products are favored. The position of the equilibrium depends on the strengths of the acids and bases. Since an acid-base reaction forms the stronger acid and base, equilibrium favors the reactants and only a small amount of product forms. H A + B A + H B + weaker acid weaker base stronger base stronger acid A larger reverse arrow means that reactants are favored. How to Predict Whether Reactants or Products are Favored in an Acid-Base Reaction HCN(g) + OH(aq) CN(aq) + H 2 O(l) acid base conjugate conjugate base acid Step [1] loss of H + gain of H + Identify the acid in the reactants and the conjugate acid in the products. 8

How to Predict Whether Reactants or Products are Favored in an Acid-Base Reaction Step [2] Determine the relative strength of the acid and the conjugate acid. From Table, HCN is a stronger acid than H 2 O. Step [3] Equilibrium favors the formation of the weaker acid. The Relative Strength of Weak Acids is Determined by the Value of the Acid Dissociation Constant, K a For the reaction where an acid (HA) dissolves in water, HA(g) + H 2 O(l) the following equilibrium constant can be written: H 3 O + (aq) + A (aq) [ ][ A ] K eq H 3O + [ HA] H 2 O [ ] HCN(g) + OH(aq) CN(aq) + H 2 O(l) stronger acid weaker acid Products are favored. [ ][ A ] K a H 3O + [ HA] K a is the acid dissociation constant. The concentration of water is not included in the equation because its concentration is essentially constant. Stronger Acids Have Larger K a Values; Weaker acids Have Smaller K a Values The Relative Strength of Weak Bases is Determined by the Value of the Base Dissociation Constant, K b For the reaction where an base (B) dissolves in water, B(aq) + H 2 O(l) BH + (aq) + OH (aq) the following equilibrium constant can be written: K eq BH + OH B [ ][ H 2 O] K b BH+ OH B [ ] K b is the base dissociation constant. The concentration of water is not included in the equation because its concentration is essentially constant. Strength of Acids Because K a values are often small and it is hard to work with small numbers, we often use the pk a value to measure the strength of an acid. pk a log K a Like with ph, the smaller the pk a value, the stronger the acid. Example: pk a of acetic acid is log(1.8 x 10 5 ) 4.74 AutoIonization of Water One water molecule acts as an acid and donates H + to the other water molecule which acts as a base and accepts the H +. H O H + H O H O H + acid loss of H + base conjugate base gain of H + concentration of H 3 O + in units of molarity (mol/l) H H O H conjugate acid + concentration of OH in units of molarity (mol/l) 9

The Ion Product Constant, K w, for Water H 2 O(l) + H 2 O(l) Experimentally it can be shown that in pure water at 25 C, [H 3 O + ] [OH ] 1.00 x 10 7 M. K w [ H 3 O + ] [ OH ] ( 1.00 10 7 M) ( 1.00 10 7 M) 1.00 10 14 M 2 K w 1.00 x 10 14 H 3 O + (aq) + OH (aq) Pure Water is Neutral: [H 3 O + ] [OH ] In pure water, the ionization of water molecules produces small but equal quantities of H 3 O + and OH ions Molar concentrations are indicated in brackets [H 3 O + ] 1.0 x 10 7 M [OH ] 1.0 x 10 7 M Acidic Solutions: [H 3 O + ] > [OH ] Adding acid to pure water Increases the [H 3 O + ] Causes the [H 3 O + ] > 1.0 x 10 7 M Decreases the [OH ] Basic Solutions: [OH ] > [H 3 O + ] Adding a base to water Increases the [OH ] Causes the [OH ] > 1.0 x 10 7 M Decreases the [H 3 O + ] Acidic, Neutral, and Basic Solutions Comparison of [H 3 O + ] and [OH ] Using the K w to Calculate [H 3 O + ] and [OH ] in Solution To calculate [OH ] when [H 3 O + ] is known: To calculate [H 3 O + ] when [OH ] is known: K w [H 3 O + ][OH ] K w [H 3 O + ][OH ] 10

Using the K w to Calculate [H 3 O + ] and [OH ] in Solution Using K w and [H 3 O + ] to Calculate [OH ] in Solution If the [H 3 O + ] in a cup of coffee is 1.0 x 10 5 M, Then the [OH ] can be calculated as follows: 1. [H 3 O + ] 1.2 x 10 5 [OH ] K w [H 3 O + ] 1.0 x 10 14 1.0 x 10 9 M 1.0 x 10 5 [ OH ] 1.00 10 14 [ H 3 O + ] 1.00 10 14 1.2 10 5 8.3 10 10 M In this cup of coffee, therefore, [H 3 O + ] > [OH ] And the solution is acidic overall. 2. [H 3 O + ] 0.27 [ OH ] 1.00 10 14 H 3 O + [ ] 1.00 10 14 3.7 10 14 M 0.27 Using K w and [OH ] to Calculate [H 3 O + ] in Solution 1. [OH ] 0.0071 [ H 3 O + ] 1.00 10 14 OH [ ] 2. [OH ] 4.2 x 10 4 [ H 3 O + ] 1.00 10 14 [ OH ] 1.00 10 14 1.4 10 12 M 0.0071 1.00 10 14 4.2 10 4 2.4 10 11 M Using K w to Calculate [H 3 O + ] and [OH ] in Solution The concentration of hydronium ions in stomach acid is 0.10 M. Calculate the concentration of hydroxide ions in stomach acid at 25 C. Strategy Use the value of K w to determine [OH - ] when [H 3 O + ] 0.10 M. Solution K w [H 3 O + ][OH - ] 1.0 10-14 at 25 C. Rearranging to solve for [OH - ], [OH - ] 1.0 10-14 [H 3 O + ] [OH - ] 1.0 10-14 0.10 [OH - ] 1.0 10-13 M Think About It Remember that the equilibrium constants are temperature dependent. The value of K w 1.0 10-14 only at 25 C. ph It s hard to work with numbers like 1.00 x 10 7 M, so chemists have adopted a shortcut notation known as ph. ph is the negative logarithm of the hydronium ion concentration. The Lower the ph, the Higher the Concentration of H 3 O + Acidic solution: ph < 7 [H 3 O + ] > 1 x 10 7 To simplify the equations, we substitute H + for H 3 O +. ph log[h + ] [H + ] 10 ph antilog( ph) Neutral solution: Basic solution: ph 7 [H 3 O + ] 1 x 10 7 ph > 7 [H 3 O + ] < 1 x 10 7 11

The ph Scale Has Values That Range From 0-14 Oven cleaners (basic), ph 14 Identify Each Solution as Acidic, Basic, Neutral HCl with a ph 1.5 >7 basic 7 neutral <7 acidic Household ammonia (basic), ph 11 Milk of Magnesia (basic), ph 10 Detergents (basic), ph 9 Blood (basic), ph 7.5 Milk (acidic), ph 6.6 Beer & wine (acidic), ph 4 Cola (acidic), ph 3 Lemon juice (acidic), ph 2 Pancreatic fluid, [H 3 O + ] 1 x 10 8 M Sprite soft drink, ph 3.0 ph 7.0 [OH ] 3 x 10 10 M [H 3 O + ] 5 x 10 12 M Hydrangea Variation in ph Values in the Human Body They produce blue flowers in acidic soil. Just add aluminum sulfate [Al 2 (SO 4 ) 3 ] or peat moss. Hydrangeas produce pink flowers in alkaline (basic) soil. Just add lime (CaO). Significant Figures in ph What is the ph of a solution with [H 3 O + ] 1.2 x 10 5 M? ph log [H 3 O + ] ph log (1.2 x 10 5 ) 2 digits ph ( 4.92) 2 decimal places ph 4.92 The number of figures to the right of the decimal in a ph is the same as the number of significant figures in the [H + ] value. Calculate the ph of the following solutions. Classify each solution as acidic or basic. 1. Enter the [H 3 O + ] value using the EE key. 2. Press the log key and change the sign. 3. Make the # of digits after the decimal point equal to the # of sig figs in the [H 3 O + ]. a. [H + ] 3.0 x 10 8 M ph log [H + ] log (3.0 x 10 8 ) 7.52, basic b. [OH ] 7.0 x 10 3 M (HINT: Use K w to obtain [H 3 O + ].) [ H 3 O + ] 1.00 10 14 [ OH ] 1.00 10 14 7.0 10 3 1.4 10 12 M ph log [H + ] log (1.4 x 10 12 ) 11.85, basic 12

Calculating ph from [H 3 O + ] Determine the ph of a solution at 25 C in which the hydronium ion concentration is (a) 3.5 10-4 M, (b) 1.7 10-7 M, and (c) 8.8 10-11 M. Strategy Given [H 3 O + ], use ph log[h 3 O + ] to solve for ph. Solution (a) ph log(3.5 10-4 ) 3.46 (b) ph log(1.7 10-7 ) 6.77 (c) ph log(8.8 10-11 ) 10.06 Think About It When a hydronium ion concentration falls between two benchmark concentrations in Table 16.4, the ph falls between the two corresponding ph values. In part (c), for example, the hydronium ion concentration (8.8 10-11 M) is greater than 1.0 10-11 M but less than 1.0 10-10 M. Therefore, we expect the ph to be between 11.00 and 10.00. Calculating [H 3 O + ] from ph If the ph of a solution is 8.50, what is the [H 3 O + ]? ph log [H 3 O + ] 8.50 log [H 3 O + ] 8.50 log [H 3 O + ] antilog ( 8.50 ) [H 3 O + ] 2 decimal places [H 3 O + ] 3.2 x 10 9 M 2 digits The solution is basic because [H 3 O + ] > 1 x 10 7 M. Summary of Steps: Calculating [H 3 O + ] From ph: [H 3 O + ] 1 x 10 ph Calculate the [H 3 O + ] for a ph value of 8.0 Calculating [H 3 O + ] From ph The ph values of specific samples of food items are listed below. Convert each value to [H + ]. a. Soft drink, ph 2.91 [H + ] 10 ph antilog( 2.91) 1.2 x 10 3 M 1. Enter the ph value, then change the sign: 8.0 2. Convert ph to concentration: Use 2 nd function key and then 10 x key or inverse key and then log key. 3. Adjust the sig figs (1 digit following the decimal point equals 1 digit in the [H 3 O + ]). b. Tomato juice, ph 4.11 [H + ] 10 ph antilog( 4.11) c. Lemon juice, ph 2.32 7.8 x 10 5 M [H + ] 10 ph antilog( 2.32) 4.8 x 10 3 M Calculating [H 3 O + ] from ph Calculate the hydronium ion concentration in a solution at 25 C in which the ph is (a) 4.76, (b) 11.95, and (c) 8.01. Strategy Given ph, use [H 3 O + ] 10 -ph to calculate [H 3 O + ]. Think About It Think About It If you use the calculated Solution hydronium ion concentrations to recalculate ph, you will get numbers (a) [Hslightly 3 O + ] 10 different -4.76 1.7 10 from those -5 M given in the problem. In part (a), for example, log(1.7 10-5 ) 4.77. The small difference between this and (b) [H4.76 3 O + ](the 10pH -11.95 given 1.1 10 in the problem) -12 M is due to a rounding error. Remember that a concentration derived from a ph with two digits to (c) [Hthe 3 O + right ] 10 of -8.01 the decimal 9.8 10point -9 M can have only two significant figures. Note also that the benchmarks can be used equally well in this circumstance. A ph between 4 and 5 corresponds to a hydronium ion concentration between 1.7 10-4 M and 1.0 10-5 M. Fun with Logarithms H 3 O + OH 1.0 10 14 Take the negative logarithm of both sides ( ) -log( 1.0 10 14 ) -log H 3 O + OH -log H 3 O + - log OH 14.00 ph + poh 14.00 13

Calculating poh from [OH ] Determine the poh of a solution at 25 C in which the hydroxide ion concentration is (a) 3.7 10-5 M, (b) 4.1 10-7 M, and (c) 8.3 10-2 M. Strategy Given [OH - ], use poh log[oh - ] to calculate poh. Solution (a) poh log(3.7 10-5 ) 4.43 (b) poh log(4.1 10-7 ) 6.39 (c) poh log(8.3 10-2 ) 1.08 Think About It Remember that the poh scale is, in essence, the reverse of the ph scale. On the poh scale, numbers below 7 indicate a basic solution, whereas number above 7 indicate an acidic solution. The poh benchmarks (abbreviated in Table 16.6) work the same way the ph benchmarks do. In part (a), for example, a hydroxide ion concentration between 1 10-4 M and 1 10-5 M corresponds to a poh between 4 and 5. Calculating [OH - ] from poh Calculate the hydroxide ion concentration in a solution at 25 C in which the poh is (a) 4.91, (b) 9.03, and (c) 10.55. Strategy Given poh, use [OH - ] 10 -poh to calculate [OH - ]. Solution (a) [OH - ] 10-4.91 1.2 10-5 M (b) [OH - ] 10-9.03 9.3 10-10 M (c) [OH - ] 10-10.55 2.8 10-11 M Think About It Use the benchmark poh values to determine whether these solutions are reasonable. In part (a), for example, the poh between 4 and 5 corresponds to [OH - ] between 1 10-4 M and 1 10-5 M. Testing the ph of a Solution ph in Solutions of Strong Acids The ph of solutions can be measured using A ph meter ph paper Indicators that have specific colors at different ph values Calculate the ph of a 0.050 M HClO 4 solution. 100% HClO 4 (aq) + H 2 O(l) H 3 O + (aq) + ClO 4 (aq) Source of hydronium ions: 1. from HClO 4 (0.050 M) 2. from autoionization of H 2 O small and can be ignored [H 3 O + ] [HClO 4 ] 0.050 M ph log(0.050) 1.30 Relating the Concentration of a Strong Acid to the ph of an Aqueous Solution Calculate the ph of an aqueous solution at 25 C that is (a) 0.035 M in HI, (b) 1.2 10-4 M in HNO 3, and (c) 6.7 10-5 M in HClO 4. Strategy HI, HNO 3, and HClO 4 are all strong acids, so the concentration of hydronium ions in each solution is the same as the stated concentration of the acid. Use Think ph About log[h It 3 O Again, + ] to calculate note that ph. when a hydronium ion concentration falls between two of the benchmark concentrations in Solution Table 16.4, the ph falls between the two corresponding ph values. (a) [H 3 In O + part ] 0.035 (b), for M example, the hydronium ion concentration of ph 1.2 10 log(0.035) -4 M is greater 1.46 than 1 10-4 M and less than 1 10-3 M. Therefore, we expect the ph to be between 4.00 and 3.00. (b) [H 3 O + ] 1.2 10-4 M ph log(1.2 10-4 ) 3.92 (c) [H 3 O + ] 6.7 10-5 M ph log(6.7 10-5 ) 4.17 Relating the Concentration of a Strong Acid to the ph of an Aqueous Solution Calculate the concentration on HCl in a solution at 25 C that has ph (a) 4.95, (b) 3.45, and (c) 2.78. Strategy Use [H 3 O + ] 10 -ph to convert from ph to [H 3 O + ]. In a strong acid solution, [H 3 O + ] is equal to the acid concentration. Solution (a) [HCl] [H 3 O + ] 10-4.95 1.1 10-5 M (b) [HCl] [H 3 O + ] 10-3.45 3.5 10-4 M (c) [HCl] [H 3 O + ] 10-2.78 1.7 10-3 M Think About It As ph decreases, acid concentration increases. 14

ph in Solutions of Weak Acids The ionization of a weak monoprotic acid HA in water is represented by: HA(aq) + H 2 O(l) K a is called the acid ionization constant. The larger the value of K a, the stronger the acid. Solution (at 25 C) H 3 O + (aq) + A (aq) + H3O A Ka HA K a ph 0.10 M HF 7.1 x 10 4 2.09 0.10 M CH 3 COOH 1.8 x 10 5 2.87 ph in Solutions of Weak Acids Calculate the ph of a 0.10 M CH 3 COOH solution. CH 3 COOH(aq) + H 2 O(l) Source of hydronium ions: 1. from CH 3 COOH (0.10 M) 2. from autoionization of H 2 O H 3 O + (aq) + CH 3 COO (aq) K a 1.8 x 10 5 K w 1.0 x 10 14 K a for CH 3 COOH is 1.8 billion times larger than K w. Thus, H 3 O + from H 2 O is so small, it can be ignored. [H 3 O + ] [CH 3 COOH] 0.10 M Initial concentration (M) ph in Solutions of Weak Acids K a H 3 O+ CH 3 COO- CH 3 COOH [ ] CH 3 COOH(aq) + H 2 O(l) ( x) ( x) 0.10 - x ( ) 1.8 10-5 H 3 O + (aq) + CH 3 COO (aq) 0.10 ~0 0 ph in Solutions of Weak Acids K a H 3 O+ CH 3 COO- CH 3 COOH [ ] x 2 0.10 - x ( x) ( x) 0.10 - x ( )» x2 ( 0.10) 1.8 10-5 ( ) 1.8 10-5 Change due to reaction (M) x x x x 2 1.8 10-6 Equilibrium concentration (M) 0.10 x x x x [H 3 O + ] 1.3 x 10 3 M ph log(1.3 x 10 3 ) 2.89 ph in Solutions of Weak Acids Calculate the ph of a 0.50 M HF solution at 25 C. HF(aq) + H 2 O(l) H 3 O + (aq) + F (aq) H O F Ka 7.110 HF + 3 4 ph in Solutions of Weak Acids HF(aq) + H 2 O(l) H 3 O + (aq) + F (aq) Initial concentration (M) 0.50 0 0 Change in concentration (M) x +x +x Equilibrium concentration (M) 0.50 x x x Initial concentration (M) HF(aq) + H 2 O(l) H 3 O + (aq) + F (aq) 0.50 0 0 + H3O F Ka HF x x 4 Ka 7.1 10 0.50 x Change in concentration (M) Equilibrium concentration (M) x 0.50 x +x +x x x Use quadratic formula to solve or Since HF is a weak acid, x could be small compared to 0.50 15

ph in Solutions of Weak Acids ph in Solutions of Weak Acids HF(aq) + H 2 O(l) H 3 O + (aq) + F (aq) Initial concentration (M) 0.50 0 0 Change in concentration (M) x +x +x Equilibrium concentration (M) 0.50 x x x HF(aq) + H 2 O(l) H 3 O + (aq) + F (aq) Initial concentration (M) 0.50 0 0 Change in concentration (M) 1.9 x 10 2 +1.9 x 10 2 +1.9 x 10 2 Equilibrium concentration (M) 0.48 1.9 x 10 2 1.9 x 10 2 x x 4 Ka 7.1 10 0.50 x simplifies x 2 (0.50)(7.1 x 10 4 ) 3.55 x 10 4 x 1.9 x 10 2 x x 4 Ka 7.1 10 0.50 ph log(0.019) 1.72 The shortcut is acceptable to use if the calculate value of x is less than 5% of the initial acid concentration 0.019M Ka 100% 3.8% 0.50M Using K a to Determine the ph of a weak Acid Solution The Ka of hypochlorous acid (HClO) is 3.5 10-8. Calculate the ph of a solution at 25 C that is 0.0075 M in HClO. Strategy Construct an equilibrium table, and express the equilibrium and concentration of each species in terms of x. Solve for x using the approximation shortcut, and evaluate whether or not the approximation is valid. Use ph log[h 3 O + ] to determine ph. HClO(aq) + H 2 O(l) H 3 O + (aq) + ClO (aq) Initial concentration (M) 0.0075 0 0 Change in concentration (M) x +x +x Equilibrium concentration (M) 0.0075 x x x Using K a to Determine the ph of a weak Acid Solution Solution These equilibrium concentrations are then substituted into the equilibrium expression to give (x)(x) K a 3.5 10 0.0075 x -8 Assuming that 0.0075 x 0.0075, Think About It We learned in Section 16.3 that the concentration of hydronium x 2 ion in 3.5 10 pure water 0.0075-8 at 25 C x 2 is 1.0 10 (3.5 10-7 -8 M, )(0.0075) yet we use 0 M as the starting concentration to solve for the ph of a weak acid. The reason Solving for this for is x, that we get the actual concentration of hydronium ion in pure water is insignificant compared 10 x to 2.625 the amount 10 produced 1.62 10 by -5 the M ionization of the weak acid. We could use the actual concentration of hydronium as the According initial concentration, to the equilibrium but doing table, so x would [H 3 O not + ]. Therefore, change the result because (x + 1.0 10-7 ) M x M. ph In solving log(1.62 10 problems -5 of ) this 4.79 type, we neglect the small concentration of H + due to the autoionization of water. ph in Solutions of Strong Bases Calculate the ph of a 0.0050 M Ca(OH) 2 solution. 100% Ca(OH) 2 (aq) Ca 2+ (aq) + 2OH (aq) [OH ] 2[Ca(OH) 2 ] 0.010 M poh log(0.010) 2.00 ph 14.00 poh 12.00 Calculations Involving [OH ], poh, and ph Calculate the poh of the following aqueous solutions at 25 C: (a) 0.013 M LiOH, (b) 0.013 M Ba(OH) 2, and (c) 9.2 10-5 M KOH. Strategy LiOH, Ba(OH) 2, and KOH are all strong bases. Use reaction stoichiometry to determine the hydroxide ion concentration and poh log[oh - ] to determine poh. Solution Think (a) About The hydroxide It These ion are concentration basic poh values, is simply which equal is what to the we concentration should expect of the base. for the Therefore, solutions [OH described - ] [LiOH] in the problem. 0.013 M. Note that while the solutions in poh parts (a) log(0.013) and (b) have 1.89 the same base concentration, they do not have the same hydroxide concentration (b) The and hydroxide therefore ion do concentration not have the same is twice poh. that of the base: Ba(OH) 2 (aq) Ba 2+ (aq) + 2OH - (aq) Therefore, [OH - ] 2[Ba(OH) 2 ] 2(0.013 M) 0.026 M. poh log(0.026) 1.59 (c) The hydroxide ion concentration is equal to the concentration of the base. Therefore, [OH - ] [KOH] 9.2 10-5 M. poh log(9.2 10-5 ) 4.04 16

Calculations Involving [OH ], poh, and ph An aqueous solution of a strong base has ph 8.15 at 25 C. Calculate the original concentration of base in the solution (a) if the base is NaOH and (b) if the base is Ba(OH) 2. Strategy Use ph + poh 14.00 to convert from ph to poh and [OH - ] 10 -poh to determine the hydroxide ion concentration. Consider the stoichiometry of dissociation Think About in each It case Alternatively, to determine we the could concentration determine of the the hydroxide base itself. ion concentration using [H Solution 3 O + ] 10-8.15 7.1 10-7 M and poh 14.00 8.15 5.85 [OH [OH - ] 1.0 10-14 - ] 7.1 10 10-5.85-9 1.4 10 1.41 10-6 -6 M (a) The Once dissociation [OH-] is of known, 1 mole the of solution NaOH produces is the same 1 mole as shown of OH-. previously. Therefore, the concentration of base is equal to the concentration of hydroxide ion. [NaOH] [OH - ] 1.41 10-6 M (b) The dissociation of 2 mole of Ba(OH) 2 produces 2 moles of OH -. Therefore, the concentration of base is only one-half the concentration of hydroxide ion. 1 [Ba(OH) 2 ] [OH - ] 7.1 10-7 M 2 ph in Solutions of Weak Bases Calculate the ph of a 0.012 M strychnine (C 21 H 22 N 2 O 2 ) solution. C 21 H 22 N 2 O 2 (aq) + H 2 O(l) Source of hydroxide ions: 1. from strychnine (0.012 M) 2. from autoionization of H 2 O C 21 H 23 N 2 O 2+ (aq) + OH (aq) K b 1.8 x 10 6 K w 1.0 x 10 14 K b for strychnine is 180 million times larger than K w. Thus, OH from H 2 O is so small, it can be ignored. [OH ] [strychnine] 0.012 M Initial concentration (M) ph in Solutions of Weak Bases K b C H N O + 21 23 2 2 OH- [ C 21 H 22 N 2 O 2 ] C 21 H 22 N 2 O 2 (aq) + H 2 O(l) ( x) ( x) 0.012 - x ( ) 1.8 10-6 C 21 H 23 N 2 O 2 + (aq) + OH (aq) 0.012 0 ~0 ph in Solutions of Weak Bases K b C H N O + 21 23 2 2 OH- [ C 21 H 22 N 2 O 2 ] x 2 0.012 - x ( x) ( x) 0.012 - x ( ) 1.8 10-6 ( )» x 2 ( 0.012) 1.8 10-6 Change due to reaction (M) x x x x 2 2.2 10-8 Equilibrium concentration (M) 0.012 x x x x [OH ] 1.5 x 10 4 M poh log(1.5 x 10 4 ) 3.83 ph 14.00 poh 10.17 Percent Ionization in Solutions of Weak Acids Using the example of a weak acid a few slides back Percent Ionization in Solutions of Weak Acids A quantitative measure of the degree of ionization is percent ionization. HF(aq) + H 2 O(l) H 3 O + (aq) + F (aq) [CH 3 COO ] 1.3 x 10 3 M [CH 3 COOH] 0.10 M Initial concentration (M) 0.50 0 0 Change in concentration (M) 1.9 x 10 2 +1.9 x 10 2 +1.9 x 10 2 amount dissociated (mol/l) Percent ionization initial concentration (mol/l) 100% ( ) 1.3 10-3 100% 1.3% 0.10 ( ) Equilibrium concentration (M) 0.48 1.9 x 10 2 1.9 x 10 2 HO 3 eq percent ionization 100% HA 0 0.019 M percent ionization 100% 3.8% 0.50 M 17

In general, the percent dissociation depends on the acid and INCREASES with increasing value of K a Also, for a given weak acid, the percent dissociation INCREASES with decreasing concentration. Percent dissociation 4.50% 4.00% 3.50% 3.00% 2.50% 2.00% 1.50% 1.00% 0.50% 0.00% 0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000 Concentration of CH 3COOH (M) Percent Ionization in Solutions of Weak Acids Calculate the percent ionization of a 1.0 M HF solution at 25 C. Initial concentration (M) 1.00 0 0 Change in concentration (M) Equilibrium concentration (M) HF(aq) + H 2 O(l) H 3 O + (aq) + F (aq) 0.027 M percent ionization 100% 2.7% 1.0 M Solution (at 25 C) 2.7 x 10 2 +2.7 x 10 2 +2.7 x 10 2 0.97 ph % ionization 0.5 M HF 1.72 3.8 1.0 M HF 1.57 2.7 2.7 x 10 2 2.7 x 10 2 Percent Ionization in Solutions of Weak Acids HF(aq) + H 2 O(l) H 3 O + (aq) + F (aq) Calculating Percent Ionization of Weak Acid Solutions Determine the ph and percent ionization for acetic acid solutions at 25 C with concentrations (a) 0.15 M, (b) 0.015 M, and (c) 0.0015 M. Strategy Using the procedure described in Worked Example 16.13, we construct an equilibrium table and for each concentration of acetic acid, we solve for the equilibrium concentration of H +. We use ph log[h 3 O + ] to find ph, and the equation below to find percent ionization. K a for acetic acid is 1.8 10-5. HO 3 eq percent ionization 100% HA 0 Solution (at 25 C) ph % ionization 0.5 M HF 1.72 3.8 1.0 M HF 1.57 2.7 Calculating Percent Ionization of Weak Acid Solutions Solution (a) CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO (aq) Initial concentration (M) 0.15 0 0 Change in concentration (M) x +x +x Equilibrium concentration (M) 0.15 x x x Solving for x gives [H 3 O + ] 0.0016 M and ph log(0.0016) 2.78. 0.0016 M percent ionization 0.15 M 100% 1.1% (b) Solving the same way as part (a) gives [H 3 O + ] 5.2 10-4 M and ph 3.28. 5.2 10-4 M percent ionization 0.015 M 100% 3.5% Calculating Percent Ionization of Weak Acid Solutions Solution (c) Solving the quadratic equation, or using successive approximation [Appendix 1] gives [H 3 O + ] 1.6 10-4 M and ph 3.78. 1.6 10-4 M percent ionization 0.0015 M 100% 11% Think About It Check your work by using the calculated value of K a to solve for the ph of a 0.10-M solution of aspirin. 18

Determine the K a of a Weak Acid Determine the K a of a Weak Acid Determine the K a of a weak acid that has a concentration of 0.25 M and a ph of 3.47 at 25 C. + H3O A Ka? HA HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) H 3 O + 10 3.47 3.39 x 10 4 M Determine the K a of a weak acid that has a concentration of 0.25 M and a ph of 3.47 at 25 C. Initial concentration (M) 0.25 0 0 Change in concentration (M) HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) 3.39 x 10 4 +3.39 x 10 4 +3.39 x 10 4 HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) Initial concentration (M) 0.25 0 0 Equilibrium concentration (M) 0.2497 3.39 x 10 4 3.39 x 10 4 Change in concentration (M) Equilibrium concentration (M) 3.39 x 10 4 +3.39 x 10 4 +3.39 x 10 4 0.2497 3.39 x 10 4 3.39 x 10 4 K a H O A + 4 3 3.39 10 2 HA Ka 4.610 0.2497 7 Using ph to Determine K a Aspirin (acetylsalicylic acid, HC 9 H 7 O 4 ) is a weak acid. It ionizes in water according to the equation HC 9 H 7 O 4 (aq) + H 2 O(l) H 3 O + (aq) + C 9 H 7 O 4- (aq) A 0.10-M aqueous solution of aspirin has a ph of 2.27 at 25 C. Determine the K a of aspirin. Strategy Determine the hydronium ion concentration from the ph. Use the hydronium ion concentration to determine the equilibrium concentrations of the other species, and plug the equilibrium concentrations into the equilibrium expressions to evaluate K a. Using ph to Determine K a Solution [H 3 O + ] 10 2.27 5.37 10-3 M To calculate K a, though, we also need the equilibrium concentrations of C 9 H 7 O - 4 and HC 9 H 7 O 4. The stoichiometry of the reaction tells us that [C 9 H 7 O 4- ] [H 3 O + ]. Furthermore, the amount of aspirin that has ionized is equal to the amount of hydronium ion in solution. Therefore, the equilibrium concentration of aspirin is (0.10 5.37 10-3 ) M 0.095 M. HC 9 H 7 O 4 (aq) + H 2 O(l) H 3 O + (aq) + C 9 H 7 O 4- (aq) Initial concentration Think About (M) It Check your 0.10 work by using the calculated 0 value of 0 K a to solve for the ph of a 0.10-M solution of aspirin. Change in concentration (M) 0.005 +5.37 10-3 +5.37 10-3 Equilibrium concentration (M) 0.095 5.37 10-3 5.37 10-3 K a [H 3O + ][C 9 H 7 O 4- ] [HC 9 H 7 O 4 ] (5.37 10-3 ) 2 3.0 10 0.095-4 The K a of aspirin is 3.0 10-4. Using K b to Calculate the ph of a Weak Base Solution What is the ph of a 0.040 M ammonia solution at 25 C. Strategy Construct an equilibrium table, and express equilibrium concentrations in terms of the unknown x. Plug these equilibrium concentrations into the equilibrium expression, and solve for x. From the value of x, determine the ph. NH 3 (aq) + H 2 O(l) NH 4+ (aq) + OH - (aq) Initial concentration (M) 0.040 0 0 Change in concentration (M) x +x +x Equilibrium concentration (M) 0.040 x x x Using K b to Calculate the ph of a Weak Base Solution Solution The equilibrium concentrations are substituted into the equilibrium expression to give K b [NH 4 + ][OH - ] (x)(x) 1.8 10 [NH 3 ] 0.040 x -5 Assuming that 0.040 x 0.040 and solving for x gives Think About It It (x)(x) is a common 1.8 10 0.040 x (x)(x) error in K b problems 0.040-5 to forget that x is hydroxide ion concentration rather than the hydronium ion concentration. Always x 2 make sure that (1.8 10-5 the ph you calculate )(0.040) 7.2 10-7 for a solution of base is a basic ph, that is, a ph greater than 7. x 7.210 7 8.5 10-4 M According to the equilibrium table, x [OH - ]. Therefore, poh log(x): log(8.5 10-4 ) 3.07 and ph 14.00 poh 14.00 3.07 10.93. The ph of a 0.040-M solution of NH 3 at 25 C is 10.93. 19

Using ph to Determine K b Caffeine, the stimulant in coffee and tea, is a weak base that ionizes in water according to the equation C 8 H 10 N 4 O 2 (aq) + H 2 O(l) HC 8 H 10 N 4 O 2+ (aq) + OH - (aq) A 0.15-M solution of caffeine at 25 C has a ph of 8.45. Determine the K b of caffeine. Strategy Use ph to determine poh, and poh to determine the hydroxide ion concentration. From the hydroxide ion concentration, use reaction stoichiometry to determine the other equilibrium concentrations and plus those concentrations into the equilibrium expression to evaluate K b. Using ph to Determine K b Solution poh 14.00 8.45 5.55; [OH - ] 10-5.55 2.82 10-6 M Based on the reaction stoichiometry, [HC 8 H 10 N 4 O 2+ ] [OH - ], and the amount of hydroxide ion in solution at equilibrium is equal to the amount of caffeine that has ionized. At equilibrium, therefore, [C 8 H 10 N 4 O 2 ] (0.15 2.82 10-6 ) M 0.15 M C 8 H 10 N 4 O 2 (aq) + H 2 O(l) HC 8 H 10 N 4 O 2+ (aq) + OH - (aq) Initial Think concentration About (M) It Check your 0.15 answer using the calculated 0 K b to 0 determine the ph ofa 0.15-M solution. Change in concentration (M) 2.82 10-6 +2.82 10-6 +2.82 10-6 Equilibrium concentration (M) 0.15 2.82 10-6 2.82 10-6 K b [HC 8H 10 N 4 O 2+ ][OH - ] (2.82 10-6 ) 2 [C 8 H 10 N 4 O 2 ] 0.15 5.3 10-11 The ph of a Polyprotic Acid Polyprotic acids can donate more than one proton. There is a K a value for each successive dissociation (,, etc.). K a1 For a typical weak polyprotic acid, K a1 >K a2 > K a3 Sulfuric acid (H 2 SO 4 ) is unique in that it is a strong acid in its first dissociation step and a weak acid in its second step. K a2 The ph of a Polyprotic Acid Diprotic and polyprotic acids undergo successive ionizations, losing one proton at a time, and each has a K a associate with it. H 2 CO 3 (aq) H + (aq) + HCO 3 (aq) HCO 3 (aq) H + (aq) + CO 3 2 (aq) K a1 > K a2 + H HCO 3 Ka1 H CO 2 3 + 2 H CO 3 Ka2 HCO 3 For a given acid, the first ionization constant is much larger than the second, and so on. Calculating Equilibrium Concentrations of All Species for a Polyprotic Acid Oxalic acid (H 2 C 2 O 4 ) is a poisonous substance used mainly as a bleaching agent. Calculate the concentrations of all species present at equilibrium in a 0.10-M solution at 25 C. Strategy Follow the same procedure for each ionization as for the determination of equilibrium concentrations for a monoprotic acid. The conjugate base resulting from the first ionization is the acid for the second ionizations, and its starting concentration is the equilibrium concentration from the first ionization. H 2 C 2 O 4 (aq) H + (aq) + HC 2 O 4 (aq) K a1 6.5 x 10 2 HC 2 O 4 (aq) H + (aq) + C 2 O 2 4 (aq) K a2 6.1 x 10 5 Construct an equilibrium table for each ionization, using x as the unknown in the first ionization and y as the unknown in the second ionization. Calculating Equilibrium Concentrations of All Species for a Polyprotic Acid Strategy H 2 C 2 O 4 (aq) H + (aq) + HC 2 O 4 (aq) Initial concentration (M) 0.10 0 0 Change in concentration (M) x +x +x Equilibrium concentration (M) 0.10 x x x The equilibrium concentration of the hydrogen oxalate (HC 2 O 4- ) after the first ionization becomes the starting concentration for the second ionization. Additionally, the equilibrium concentration of H + is the starting concentration for the second ionization. HC 2 O 4 (aq) H + (aq) + C 2 O 2 4 (aq) Initial concentration (M) x x 0 Change in concentration (M) y +y +y Equilibrium concentration (M) x y x + y y 20

Calculating Equilibrium Concentrations of All Species for a Polyprotic Acid Solution [H + ][HC 2 O 4- ] K a1 [H 2 C 2 O 4 ] 6.5 10-2 x 2 0.10 x Applying the approximation and neglecting x in the denominator of the expression gives 6.5 10-2 x 2 0.10 x 2 6.5 10-3 x 8.1 10-2 M Testing the approximation, 8.1 10-2 M 100% 81% 0.10 M Clearly the approximation is not valid, so we must solve the following quadratic equation: x 2 + 6.5 10-3 x 6.5 10-3 0 Calculating Equilibrium Concentrations of All Species for a Polyprotic Acid Solution The result x 0.054 M. Thus, after the first ionization, the concentrations of species in solution are [H + ] 0.054 M [HC 2 O 4- ] 0.054 M [H 2 C 2 O 4 ] (0.10 0.054) M 0.046 M Rewriting the equilibrium table for the second ionization, using the calculated value of x, gives the following: HC 2 O 4 (aq) H + (aq) + C 2 O 2 4 (aq) Initial concentration (M) 0.054 0.054 0 Change in concentration (M) y +y +y Equilibrium concentration (M) 0.054 y 0.054 + y y K a2 [H + ][C 2 O 4 2- ] [HC 2 O 4- ] 6.1 10-5 (0.054 + y)(y) 0.054 y Calculating Equilibrium Concentrations of All Species for a Polyprotic Acid Solution Assuming that y is very small and applying the approximations 0.054 + y 0.054 and 0.054 y 0.054 gives (0.054)(y) 0.054 y 6.1 10-5 We must test the approximation as follows to see if it is valid: Think About It Note that the second ionization did not contribute significantly to the H+ 6.1 10 concentration. -5 M Therefore, we could determine the ph of this solution 100% 0.11% 0.054 by considering M only the first ionization. This is true in general for polyprotic acids where K This time, because the ionization constant is much smaller, a1 is at least the approximation is 1000 K valid. At equilibrium, a2. [Note that it is necessary to consider the second the concentrations of all species are ionization to determine the concentration of oxalate ion (C 2 O 2-4 ).] [H 2 C 2 O 4 ] 0.046 M [HC 2 O 4- ] (0.054 6.1 10-5 ) 0.054 M [H + ] (0.054 + 6.1 10-5 ) 0.054 M [C 2 O 4 2- ] 6.1 10-5 M K a and K b Relationship NH + 4 (aq) + H 2 O(l) H 3 O + (aq) + NH 3 (aq) K a H 3O + [ NH 3] 5.6 10-10 + NH 4 acid NH 3 (aq) + H 2 O(l) NH + 4 (aq) + OH (aq) K b NH + 4 OH- 1.8 10-5 [ NH 3 ] base 2H 2 O(l) H 3 O + (aq) + OH (aq) K w H 3 O + OH- 1.0 10-14 For any given acid base pair, K a K b K w Determining Ionization Constants for Conjugates Determine (a) K b of the acetate ion (CH 3 COO - ), (b) K a of the methylammonium ion (CH 3 NH 3+ ), (c) K b of the fluoride (F - ), and (d) K a of the ammonium ion (NH 4+ ). Strategy Each species listed is either a conjugate base or a conjugate acid. Determine the identity of the acid corresponding to each conjugate base and the identity of the base corresponding to each conjugate acid; then, consult Table 16.7 and 16.8 for their ionization constants. Use the tabulated ionization constants and K w K a K b to calculate each indicated K value. K K w a K b and K K w b K a Solution (a) A K b value is requested, indicating that the acetate ion is a conjugate base. To identify the corresponding Brønsted acid, add a proton to the formula to get CH 3 COOH (acetic acid). The K a of acetic acid is 1.8 10-5. Conjugate base CH 3 COO - 1.0 10 : K b -14 5.6 10 1.8 10-5 -10 Determining Ionization Constants for Conjugates Solution (b) A K a value is requested, indicating that the methylammonium ion is a conjugate acid. Determine the identity of the corresponding Brønsted base by removing a proton from the formula to get CH 3 NH 2 (methylamine). The K b of methylamine is 4.4 10-4. 1.0 10 Conjugate acid CH 3 NH 3+ : K a -14 2.3 10 4.4 10-11 Think About It Because the conjugates -4 of weak acids and bases (c) F - is have the conjugate ionization base constants, of HF; salts Ka containing 7.1 10-4 these. ions have an effect on the ph of a solution. In Section Conjugate base F - 1.0 10 : K b -14 16.10 we will use the ionization constants of conjugate acids and 1.4 10 7.1 10-4 conjugates bases -11 to calculate ph for solutions containing dissolved salts. (d) NH + 4 is the conjugate acid of NH 3 ; K b 1.8 10-5. Conjugate acid NH 4+ : K a 1.0 10-14 1.8 10-5 5.6 10-10 21

The Acidity and Basicity of Salt Solutions When an acid neutralizes a base, an ionic compound called a salt is formed (e.g. NaCl, CaBr 2, Mg(NO 3 ) 2 ) A salt derived from a strong acid and a strong base forms a neutral solution (ph 7). Na + from NaOH strong base NaCl Cl from HCl strong acid Salt solutions can be neutral, acidic, or basic, depending on the acids and bases that reacted to form them. For the salt M + A : The cation M + comes from the base. The anion A comes from the acid HA. A salt derived from a strong base and a weak acid forms a basic solution (ph > 7). A salt derived from a weak base and a strong acid forms an acidic solution (ph < 7). Na + from NaOH strong base NH 4 + from NH 3 weak base NaHCO 3 NH 4 Cl HCO 3 from H 2 CO 3 weak acid Cl from HCl strong acid The Acidity and Basicity of Salt Solutions The ion derived from the stronger acid or base determines whether the solution is acidic or basic. ph of an Acidic Salt Solution Calculate the ph of a 0.10 M ammonium chloride, NH 4 Cl, solution. NH 4 Cl(s) NH 4+ (aq) + Cl (aq) NH 4+ (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) acid base base acid [ ] H 3 O + K a NH 3 + NH 4 A value for the K a of NH 4 + is generally not published. ph of an Acidic Salt Solution NH 4 + is the conjugate acid of NH 3 (a base). For any given acid base pair, ph of an Acidic Salt Solution [ ] H 3 O + K a NH 3 + NH 4 ( x) ( x) 5.6 10-10 0.10 - x ( ) K a K b K w NH 4 + (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) K a K w 1.0 10-14 5.6 10-10 -5 K b 1.8 10 Initial concentration (M) Change due to reaction (M) 0.10 0 ~0 x x x Equilibrium concentration (M) 0.10 x x x 22

ph of an Acidic Salt Solution [ ] H 3 O + K a NH 3 + NH 4 x 2 0.10 - x ( )» x2 ( 0.10) ( x) ( x) 5.6 10-10 0.10 - x ( ) 5.6 10-10 ph of an Basic Salt Solution Sodium benzoate is used as a food preservative. Calculate the ph of a 0.015 M sodium benzoate solution. NaOH(aq) + HC 7 H 5 O 2 (aq) NaC 7 H 5 O 2 (aq) Strong base Weak acid Basic solution x 2 5.6 10-11 x [H 3 O + ] 7.5 x 10 6 M ph log(7.5 x 10 6 ) 5.12 ph of an Basic Salt Solution Sodium benzoate is used as a food preservative. Calculate the ph of a 0.015 M sodium benzoate solution. NaC 7 H 5 O 2 (s) Na + (aq) + C 7 H 5 O 2 (aq) ph of an Basic Salt Solution [ ] OH - K b HC 7H 5 O 2 - C 7 H 5 O 2 ( x) ( x) 0.015- x ( ) 1.6 10-10 C 7 H 5 O 2 (aq) + H 2 O(l) HC 7 H 5 O 2 (aq) + OH (aq) C 7 H 5 O 2 (aq) + H 2 O(l) base acid HC 7 H 5 O 2 (aq) + OH (aq) acid base Initial concentration (M) 0.015 0 ~0 [ ] OH - K b HC 7H 5 O 2 - C 7 H 5 O 2 K w K a 1.0 10-14 1.6 10-10 -5 6.3 10 Change due to reaction (M) Equilibrium concentration (M) x 0.015 x x x x x ph of an Basic Salt Solution [ ] OH - K b HC 7H 5 O 2 - C 7 H 5 O 2 ( x) ( x) 0.015- x ( ) 1.6 10-10 ph of a Salt Solution Resulting from a Weak Acid and a Weak Base The calculations are quite complex; we won t do them. x 2 0.015- x ( )» x 2 ( 0.015) x 2 2.4 10-12 1.6 10-10 Weak acid + Weak Base K a > K b K b > K a Acidic, Basic, or Neutral solution ph < 7 (acidic) ph > 7 (basic) x [OH ] 1.5 x 10 6 M poh log(1.5 x 10 6 ) 5.82 ph 14.00 poh 8.18 K a K b ph 7 (neutral) 23

Calculating the ph of a Basic Salt Solution Calculate the ph of a 0.10-M solution of sodium fluoride (NaF) at 25 C. Strategy A solution of NaF contains Na + ions and F - ions. The F - ion is the conjugate base of the weak acid, HF. Use the K a value for HF (7.1 10-4 ) and K w K b K a to determine K b for F - : K 1.0 10 K -14 w 1.4 10-11 b K 7.1 10-4 a Then, solve this ph problem like any equilibrium problem, using an equilibrium table. F - (aq) + H 2 O(l) HF(aq) + OH - (aq) K b [HF][OH - ] [F - ] F - (aq) + H 2 O(l) HF(aq) + OH - (aq) Initial concentration (M) 0.10 0 0 Change in concentration (M) x +x +x Equilibrium concentration (M) 0.10 x x x Calculating the ph of a Basic Salt Solution Solution Substituting the equilibrium concentrations into the equilibrium expression and using the shortcut to solve x, we get 1.4 10-11 x 2 0.10 x x 2 0.10 11 x (1.4 10 )(0.10) 1.2 10-6 M Think About It It s easy to mix up ph and poh in this type of problem. Always make a qualitative prediction According to our equilibrium table, x [OH - regarding the ph of a ]. In this case, the autoionization of salt solution first, and then check to make sure that your calculated water makes a significant contribution to the hydroxide ion concentration so the ph agrees with your prediction. In this case, total concentration will be the sum of 1.2 10-6 we would predict a M (from the ionization of F - ) and basic 1.0 10-7 ph because the anion in the salt (F M (from the autoionization of water). - ) is the conjugate base of a Therefore, we calculate the poh weak acid (HF). The calculated ph, 8.05, is indeed basic. first as poh log(1.2 10-6 + 1.0 10-7 ) 5.95 and then the ph, ph 14.00 poh 14.00 5.95 8.05 The ph of a 0.10-M solution of NaF at 25 C is 8.05. Calculating the ph of an Acidic Salt Solution Calculate the ph of a 0.10-M solution of ammonium chloride (NH 4 Cl) at 25 C. Strategy A solution of NH 4 Cl contains NH + 4 ions and Cl - ions. The NH + 4 ion is the conjugate acid of the weak base, NH 3. Use the K b value for NH 3 (1.8 10-5 ) and K w K b K a to determine K a for F - : K 1.0 10 K -14 w 5.6 10-10 a K 1.8 10-5 b Again, we write the balanced chemical equation and the equilibrium expression: NH 4 + (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) K b [NH 3 ][H 3 O + ] [NH 4 + ] NH 4 + (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) Initial concentration (M) 0.10 0 0 Change in concentration (M) x +x +x Calculating the ph of an Acidic Salt Solution Solution Substituting the equilibrium concentrations into the equilibrium expression and using the shortcut to solve x, we get 5.6 10-10 x 2 0.10 x x x 2 0.10 10 (5.610 )(0.10) 7.5 10-6 M According to our equilibrium table, x [H 3 O + ]. The ph can be calculated as follows: ph log(7.5 10-6 ) 5.12 The ph of a 0.10-M solution of ammonium chloride (at 25 C) is 5.12. Think About It In this case, we would predict an acidic ph because the cation in the salt (NH 4+ ) is the conjugate acid of a weak base (NH 3 ). The calculated ph is acidic. Equilibrium concentration (M) 0.10 x x x Molecular Structure and Acid Strength The strength of an acid is measured by its tendency to ionize. HA H + + A There are two factors that determine the strength of an acid: Molecular Structure and Acid Strength The electrostatic potential maps show that all these molecules are polar, with the halogen atom being electron rich (red) and the H atom being electron poor (blue). 1. The strength of the H A bond. 2. The polarity of the H A bond. δ+ δ H A 24

Molecular Structure and Acid Strength In general, for binary acids of elements in the same group of the periodic table, the H A bond strength is the most important determinant of acidity. Molecular Structure and Acid Strength For binary acids in the same row of the periodic table, changes in the H A bond strength are smaller and the polarity of the H A bond is the most important determinant of acidity. Only HF is a weak acid. Molecular Structure and Acid Strength Oxoacids (a.k.a oxyacids) have the general formula H n YO m, where An oxoacid contains H, O, and a central nonmetal atom. Y is a nonmetallic atom, such as C, N, S, or Cl n and m are integers The atom Y is always bonded to one or more hydroxyl (OH) groups and can be bonded, in addition, to one or more oxygen atoms. Molecular Structure and Acid Strength Because the dissociation of an oxoacid requires breaking on O H bond, any factor that weakens the O H bond or increases its polarity increases the strength of the acid. Molecular Structure and Acid Strength For oxoacids that contain the same number of OH groups and the same number of O atoms, acid strength increases as the electronegativity of Y increases. Y O H + H 2 O H 3 O + + Y O 25

Molecular Structure and Acid Strength For oxoacids that contain the same atom Y but different numbers of oxygen atoms, acid strength increases as the number of oxygen atoms increases. Comparison of Acid Strengths Based on Molecular Structure Predict the relative strengths of the oxoacids in each of the following groups: (a) HClO, HBrO, and HIO; (b) HNO 3 and HNO 2. Strategy In each group, compare the electronegativies or oxidation numbers of the central atoms to determine which O H bonds are the most polar. The more polar the O H bond, the more readily it is broken and the stronger the acid. Solution Think (a) About In a group It Four with of different the strong central acids atoms, are oxoacids: we must HNO compare 3, electronegativities. HClO 4, HClOThe 3, and electronegativities H 2 SO 4. of the central atoms in this group decrease as follows: Cl > Br > I. Acid strength decreases as follows: HClO > HBrO > HIO (b) These two acids have the same central atom but differ in the number of oxygen atoms. In a group such as this, the greater the number of attached oxygen atoms, the higher the oxidation number and the stronger the acid. HNO 3 is stronger than HNO 2. Identify the stronger acid in each of the following pairs: Lewis acid: Lewis Acids and Bases an electron pair acceptor a. H 2 S or H 2 Se H 2 Se Lewis base: an electron pair donor b. HI or H 2 Te HI empty unhybridized 2p z orbital c. HNO 2 or HNO 3 HNO 3 Boron trifluoride a Lewis acid Ammonia, a Lewis base A coordinate covalent bond Lewis Acids and Bases Identify the Lewis acid and the Lewis base in the following reaction: CO 2 + OH HCO 3 d - d + d - O C O O + OH O C OH Lewis acid Lewis base 26